# Sober subspaces and the Skula topology

It often happens that one wishes to show that a certain subspace A of a given sober space X is sober. The following is a pearl due to Keimel and Lawson [1, Corollary 3.5], which was mentioned to me by Zhenchao Lyu in July:

Given a sober space X, the subsets A of X that are sober as subspaces are exactly its Skula-closed subsets.

My goal today is to comment on that, and to (re)prove it.

## In search of sober subspaces

It is not hard to show that if A is a closed subspace of a sober space X, then it is sober. This is also true if A is open.

I have regularly used the following practical result (Lemma 8.4.12 in the book): every subspace A of a sober space X that arises as the equalizer [g1=g2] of two continuous maps g1=g2: XY, where Y is any T0 space, is sober. (By the way, Y has to be T0. This is one of the mistakes in the book.)

In particular, if A is open in X, then it arises as the equalizer of its characteristic map χA and the constant map equal to 1 from X to Sierpiński space S (the space {0, 1} with the Alexandroff topology of 0<1). And if A is closed, then it arises as the equalizer of the characteristic map χU of its complement U and the constant map equal to 0 from X to S.

As another application of Lemma 8.4.12 of the book, one can show that every Π02 subspace of a sober space X is sober again. A Π02 subset is the intersection of countably many sets Un Vn, where each Un and each Vn is open in X. (UnVn is the subset of points x such that if x is in Un then x is in Vn.) That result was used several times by Matthew de Brecht in the study of quasi-Polish spaces.

How do you prove that? Well, take Y to be the product of countably many copies of S. Define g1(x) as the tuple whose nth entry is χUn (x), and g2(x) as the tuple whose nth entry is χUnVn (x). Then the equalizer of g1 and of g2 is the set of points x such that x is in Un if and only if it is in UnVn, and that is exactly our Π02 subset. (Alternatively, let Y be the powerset of N, with its Scott topology, define g1(x) as the set of natural numbers n such that x is in Un, and g2(x) as the set of natural numbers n such that x is in UnVn.)

In that argument, note that the fact that we are taking a countable intersection of sets UnVn (so-called UCO sets) is entirely irrelevant. Any intersection of UCO subsets of a sober space X, of whatever cardinality, is sober.

## The Skula topology

The Skula topology (also called the strong topology) on a topological space X is the topology generated by the open subsets and the closed subsets of X. It is equivalent to define it as the topology generated by the open subsets and the downwards-closed subsets of X, because every downwards-closed subset of X is a union of closed sets (namely the downward closures of single points).

This is a rather remarkable topology. For example, a theorem by R.-E. Hoffmann states that X is Skula-compact (i.e., compact in its Skula topology) if and only if X is sober Noetherian (Exercise 9.7.16 in the book; note that the Skula topology is always T2.)

A subset of X is Skula-closed (i.e., closed in the Skula topology) if and only if it is a (generally infinite) intersection of unions VC of an open subset V and a closed subset C of X. If you write U for the complement of C, you will realize that such a union VC is nothing but the UCO subset UV. But recall that any intersection of UCO subsets of a sober space is sober in the subspace topology of X. Thus we obtain one half of Keimel and Lawson’s Lemma: every Skula-closed subset A of a sober space X is sober in the subspace topology from X.

In the converse direction, let A be a sober subspace of X, and assume that A is not Skula-closed. Hence its complement is not Skula-open, and that implies that there is a point x in XA whose Skula-open neighborhoods all intersect A. (Reason by contradiction: otherwise every point in XA would have a Skula-open neighborhood included in XA.)

In particular, for every open neighborhood U of x (in the original topology of X), U ∩ ↓x is a Skula-open neighborhood of x, so it must intersect A. Said in another way: (*) every open neighborhood U of x intersects ↓xA.

Note that ↓xA is a closed subset of A. We claim that it is irreducible in A.

Using (*) with U=X, we obtain that ↓xA is non-empty. Given two open subsets U and V of X that intersect ↓xA, both U and V must contain x, so UV also contains x, and using (*) we obtain that UV intersects ↓xA. It follows that ↓xA is irreducible closed in A.

Since A is sober, ↓xA is the closure of some unique point y in A.

In particular, y is in ↓xA, so yx. Since x is in XA, and y is in A, we have xy, so x is not below y. We use (*) with U equal to the complement of ↓y: there is a point in ↓xA and in U; since ↓xA is the closure of y in A, and since an open set intersects the closure of a set B if and only if it intersects B itself, y must be in U—but that contradicts the definition of U.

We have proved the promised statement:

Proposition [1, Corollary 3.5]. Given a sober space X, the subsets A of X that are sober as subspaces are exactly its Skula-closed subsets.

This also shows that Lemma 8.4.12 of the book is in a sense optimal: given any sober subspace A of a sober space X, A is the equalizer [g1=g2] of two continuous maps g1=g2: XY. We can even take Y to be a power of S (equivalently, a powerset of some set, in its Scott topology). Indeed, A is Skula-closed, hence an intersection of some family of UCO subsets, and we have already seen that any such intersection can be realized as such an equalizer.

## Sobrifications

Corollary 3.5 of  is—as the name indicates—a corollary of a more general result (I am using the notion of sobrification from the book):

Theorem [1, Proposition 3.4]. Let X be a sober space, and A be a subset of X. The sobrification S(A) of the subspace A is homeomorphic to its Skula-closure in X.

Proof. We look at the inclusion map i from A into its Skula-closure cls(A). This is a continuous map, and since cls(A) is sober, i has a unique continuous extension j from S(A) to cls(A)—namely, j o ηA = i, where the embedding ηA : AS(A) maps x to its downward closure in A, namely ↓xA.

Note that, for every open subset of S(A), namely for every set of the form ♢(UA) where U is open in X, ηA-1(♢(UA))=UA. Since i-1(U ∩ cls(A))=UA, and recalling that ηA-1 is a frame isomorphism, it follows that j-1(U ∩ cls(A))=♢(UA) for every open subset U of X.

In the converse direction, given any point x in cls(A), we let f(x)=↓xA. For every open subset U of X, f(x) intersects U if and only if U intersects ↓xA, if and only if A intersects the Skula-open set ↓xU, if and only if cls(A) intersects ↓xU, if and only if x is in U (recall that x is in cls(A): if cls(A) intersects ↓xU, say at y, then yx and y is in U, so x is in U, and conversely if x is in U, then cls(A) intersects ↓xU at x). To sum up: (**) for every open subset U of X, for every x in cls(A), f(x) intersects U if and only if x is in U.

Let us fix x in cls(A). Using (**) with U empty, we obtain that f(x) is not empty; given any two open sets U and V that each intersect f(x), by (**) we obtain that both U and V contain x, so UV also contains x, and using (**) again, f(x) intersects UV. This shows that f(x) is irreducible (closed) in A, hence an element of S(A).

We can now rephrase (**) as: for every open subset U of X, f-1(♢(UA))=U ∩ cls(A). In particular, f is continuous. It also follows that f-1 is a frame homomorphism that is inverse to the frame homomorphism j-1. Since f and j are continuous maps between sober spaces, they are uniquely determined by the frame homomorphisms f-1 and j-1. It follows that f and j are mutual inverses. ☐

1. Klaus Keimel and Jimmie D. Lawson.  D-completions and the d-topology.  Annals of Pure and Applied Logic 159(3), June 2009, pages 292-306.

Jean Goubault-Larrecq 